# How many prime factors are obtained? x^7 + c^3x^4 - c^4x^3-c^7

## ${x}^{7} + {c}^{3} {x}^{4} - {c}^{4} {x}^{3} - {c}^{7}$

May 9, 2018

${x}^{7} + {c}^{3} {x}^{4} - {c}^{4} {x}^{3} - {c}^{7} = \left({x}^{2} - c x + {c}^{2}\right) \left({x}^{2} + {c}^{2}\right) {\left(x + c\right)}^{2} \left(x - c\right)$

#### Explanation:

${x}^{7} + {c}^{3} {x}^{4} - {c}^{4} {x}^{3} - {c}^{7}$

Not so easy this one.

First note that if $x = c$ then the expression yields zero.

${c}^{7} + {c}^{3} {c}^{4} - {c}^{4} {c}^{3} - {c}^{7} = {c}^{7} + {c}^{7} - {c}^{7} - {c}^{7} = 0$

This means that $\left(x - c\right)$ is a factor of ${x}^{7} + {c}^{3} {x}^{4} - {c}^{4} {x}^{3} - {c}^{7}$.

Let's do the long division and see what remains.

${x}^{7} + {c}^{3} {x}^{4} - {c}^{4} {x}^{3} - {c}^{7}$

$= {x}^{7} - c {x}^{6} + c {x}^{6} - {c}^{2} {x}^{5} + {c}^{2} {x}^{5} - {c}^{3} {x}^{4} + 2 {c}^{3} {x}^{4} - 2 {c}^{4} {x}^{3} + {c}^{4} {x}^{3} - {c}^{5} {x}^{2} + {c}^{5} {x}^{2} - {c}^{6} x + {c}^{6} x - {c}^{7}$

$= {x}^{6} \left(x - c\right) + c {x}^{5} \left(x - c\right) + {c}^{2} {x}^{4} \left(x - c\right) + 2 {c}^{3} {x}^{3} \left(x - c\right) + {c}^{4} {x}^{2} \left(x - c\right) + {c}^{5} x \left(x - c\right) + {c}^{6} \left(x - c\right)$

$= \left({x}^{6} + c {x}^{5} + {c}^{2} {x}^{4} + 2 {c}^{3} {x}^{3} + {c}^{4} {x}^{2} + {c}^{5} x + {c}^{6}\right) \left(x - c\right)$

Now we notice that when $x = - c$,

${x}^{6} + c {x}^{5} + {c}^{2} {x}^{4} + 2 {c}^{3} {x}^{3} + {c}^{4} {x}^{2} + {c}^{5} x + {c}^{6} = {c}^{6} - {c}^{6} + {c}^{6} - 2 {c}^{6} + {c}^{6} - {c}^{6} + {c}^{6} = 0$

Let's divide ${x}^{6} + c {x}^{5} + {c}^{2} {x}^{4} + 2 {c}^{3} {x}^{3} + {c}^{4} {x}^{2} + {c}^{5} x + {c}^{6}$ by $\left(x + c\right)$.

${x}^{6} + c {x}^{5} + {c}^{2} {x}^{4} + 2 {c}^{3} {x}^{3} + {c}^{4} {x}^{2} + {c}^{5} x + {c}^{6}$

$= {x}^{6} + c {x}^{5} + {c}^{2} {x}^{4} + {c}^{3} {x}^{3} + {c}^{3} {x}^{3} + {c}^{4} {x}^{2} + {c}^{5} x + {c}^{6}$

$= {x}^{5} \left(x + c\right) + {c}^{2} {x}^{3} \left(x + c\right) + {c}^{3} {x}^{2} \left(x + c\right) + {c}^{5} \left(x + c\right)$

$= \left({x}^{5} + {c}^{2} {x}^{3} + {c}^{3} {x}^{2} + {c}^{5}\right) \left(x + c\right)$

Next notice that when $x = - c$

${x}^{5} + {c}^{2} {x}^{3} + {c}^{3} {x}^{2} + {c}^{5} = - {c}^{5} - {c}^{5} + {c}^{5} + {c}^{5} = 0$

So lets divide ${x}^{5} + {c}^{2} {x}^{3} + {c}^{3} {x}^{2} + {c}^{5}$ by $\left(x + c\right)$.

${x}^{5} + {c}^{2} {x}^{3} + {c}^{3} {x}^{2} + {c}^{5}$

$= {x}^{5} + c {x}^{4} - c {x}^{4} - {c}^{2} {x}^{3} + 2 {c}^{2} {x}^{3} + 2 {c}^{3} {x}^{2} - {c}^{3} {x}^{2} - {c}^{4} x + {c}^{4} x + {c}^{5}$

$= {x}^{4} \left(x + c\right) - c {x}^{3} \left(x + c\right) + 2 {c}^{2} {x}^{2} \left(x + c\right) - {c}^{3} x \left(x + c\right) + {c}^{4} \left(x + c\right)$

$= \left({x}^{4} - c {x}^{3} + 2 {c}^{2} {x}^{2} - {c}^{3} x + {c}^{4}\right) \left(x + c\right)$

Now we are forced to factor ${x}^{4} - c {x}^{3} + 2 {c}^{2} {x}^{2} - {c}^{3} x + {c}^{4}$.

A little "inspired grouping" gets us over this hurdle.

${x}^{4} - c {x}^{3} + 2 {c}^{2} {x}^{2} - {c}^{3} x + {c}^{4}$.

Split the third term.

${x}^{4} - c {x}^{3} + {c}^{2} {x}^{2} + {c}^{2} {x}^{2} - {c}^{3} x + {c}^{4}$

Commute the 3rd term.

${x}^{4} + {c}^{2} {x}^{2} - c {x}^{3} - {c}^{3} x + {c}^{2} {x}^{2} + {c}^{4}$

Group and factor.

${x}^{2} \left({x}^{2} + {c}^{2}\right) - c x \left({x}^{2} + {c}^{2}\right) + {c}^{2} \left({x}^{2} + {c}^{2}\right)$

$\left({x}^{2} - c x + {c}^{2}\right) \left({x}^{2} + {c}^{2}\right)$

If $c$ and $x$ are both real numbers, then we cannot factor any further. Note that the only way for ${x}^{2} + {c}^{2} = 0$ is if they are both zero or if either ${x}^{2}$ or ${c}^{2}$ are less than zero. The only way for to have a squared number be less than zero is if it is imaginary. Also note that by using the quadratic formula, we can find $x$'s that satisfy ${x}^{2} - c x + {c}^{2} = 0$.

$x = \frac{c \pm \sqrt{{c}^{2} - 4 {c}^{2}}}{2} = \frac{c \pm c \sqrt{- 3}}{2}$

So both roots for this expression MUST be imaginary regardless of whether $c$ is imaginary or not.

So we have the factored form with the following logic:

${x}^{7} + {c}^{3} {x}^{4} - {c}^{4} {x}^{3} - {c}^{7}$

$= \left({x}^{6} + c {x}^{5} + {c}^{2} {x}^{4} + 2 {c}^{3} {x}^{3} + {c}^{4} {x}^{2} + {c}^{5} x + {c}^{6}\right) \left(x - c\right)$

$= \left({x}^{5} + {c}^{2} {x}^{3} + {c}^{3} {x}^{2} + {c}^{5}\right) \left(x + c\right) \left(x - c\right)$

$= \left({x}^{4} - c {x}^{3} + 2 {c}^{2} {x}^{2} - {c}^{3} x + {c}^{4}\right) \left(x + c\right) \left(x + c\right) \left(x - c\right)$

$= \left({x}^{2} - c x + {c}^{2}\right) \left({x}^{2} + {c}^{2}\right) \left(x + c\right) \left(x + c\right) \left(x - c\right)$

$= \left({x}^{2} - c x + {c}^{2}\right) \left({x}^{2} + {c}^{2}\right) {\left(x + c\right)}^{2} \left(x - c\right)$

Yikes! Your teacher sure is mean!